4 Answers
4

If two lines $Ax+By+C=0$ and $Dx+Ey+F=0$ are parallel, then they have no points of intersection (or infinitely many if you allow a line to be parallel to itself). In either case, the system of equations
$$\begin{cases}
Ax+By+C=0,\\
Dx+Ey+F=0
\end{cases}$$
will not have a single, unique solution. The system of equations is equivalent to
$$\begin{bmatrix}
A & B\\
D & E
\end{bmatrix}\cdot\begin{bmatrix}
x\\
y
\end{bmatrix}+\begin{bmatrix}
C\\
F
\end{bmatrix}=0$$
or
$$\begin{bmatrix}
A & B\\
D & E
\end{bmatrix}\cdot\begin{bmatrix}
x\\
y
\end{bmatrix}=\begin{bmatrix}
-C\\
-F
\end{bmatrix}.$$
This matrix equation has a single unique solution if and only if the coefficient matrix $\begin{bmatrix}
A & B\\
D & E
\end{bmatrix}$ is invertible, which is true if and only if its determinant is nonzero.

So, if the lines are parallel, then
$$\begin{vmatrix}
A & B\\
D & E
\end{vmatrix}=AE-BD=0$$
or
$$\frac{A}{D}=\frac{B}{E}$$
which means that $\frac{A}{D}(Dx+Ey+F)=Ax+By+\frac{FA}{D}=0$, so $Dx+Ey+F=0$ can be written in the form $Ax+By+k=0$.

edit: from Arturo Magidin's comment below, to handle the cases where $D=0$ or $E=0$ (in which $A/D = B/E$ makes no sense):

If $D=0$ then $A=0$ or $E=0$, but you cannot have both $D$ and $E$ zero. So $D=0$ implies $A=0$, and $B\neq 0\neq E$. Similarly, if $E=0$, then $B=0$ and $A\neq 0 \neq D$. And you get the similar conclusion without forgetting about horizontal and vertical lines.

You also need to allow for the possibility that $D=0$ or $E=0$, so that $A/D = B/E$ make no sense. If $D=0$ then $A=0$ or $E=0$, but you canno thave both $D$ and $E$ zero. So $D=0$ implies $A=0$, and $B\neq 0\neq E$. Similarly, if $E=0$, then $B=0$ and $A\neq 0 \neq D$. And you get the similar conclusion without forgetting about horizontal and vertical lines.
–
Arturo MagidinSep 18 '10 at 20:40

Examples don't show it in any case. You cannot prove that something holds for infinitely many things by doing nothing except checking finitely many examples.

What you need to do is remember that two lines are parallel if and only if they have the same slope or they are both vertical (of the form $x=k$ for some constant $k$). Then figure out what the slope of an arbitrary line $\alpha x+\beta y+\gamma=0$ would be in terms of $\alpha$, $\beta$, and $\gamma$ (careful when dividing by constants to consider what happens if the constant you are dividing by is equal to zero). Then see what this means if you want to find all lines $\alpha x+\beta y+\gamma = 0$ that are parallel to a given line $Ax+By+c=0$.

I think "You cannot prove that something holds for infinitely many things by checking finitely many examples" is too strong a statement to be true. Certainly though, you can't prove this theorem by picking a set of 20 different values of A,B,c,k and checking.
–
anonSep 18 '10 at 19:40

Fair enough; should be "only by checking finitely many examples". You could do it if you also reduce the problem to a finite set of potential counterexamples in addition to checking those exemples.
–
Arturo MagidinSep 18 '10 at 19:46

It is a preliminary and useful exercise to show that the line $Ax+By+C=0$ and the vector $v=(A,B)$ are perpendicular (one says that $v$ is the normal vector of the line). With these, all amounts to observe that two lines are parallel if and only if their normal vectors are parallel.